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            	<h1 id="-">斐波那契数列</h1>
<p>难度简单141收藏分享切换为英文接收动态反馈</p>
<p>写一个函数，输入 <code>n</code> ，求斐波那契（Fibonacci）数列的第 <code>n</code> 项（即 <code>F(N)</code>）。斐波那契数列的定义如下：</p>
<pre><code>F(0) = 0,   F(1) = 1
F(N) = F(N - 1) + F(N - 2), 其中 N &gt; 1.
</code></pre><p>斐波那契数列由 0 和 1 开始，之后的斐波那契数就是由之前的两数相加而得出。</p>
<p>答案需要取模 1e9+7（1000000007），如计算初始结果为：1000000008，请返回 1。</p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：n = 2
输出：1
</code></pre><p><strong>示例 2：</strong></p>
<pre><code>输入：n = 5
输出：5
</code></pre><hr>
<p>思路：总结规律，已知当n &gt; 1时，f(n) = f(n - 1) + f(n - 2)</p>
<pre><code class="lang-java">class Solution {
    public int fib(int n) {
        if (n == 0 || n == 1)
            return n;
        //a为 f(n - 1) ， b为f(n - 2)
        //范围1至n
        int a = 1, b = 0;
        for (int i = 1; i &lt; n; i++) {
            a = a + b;
            b = a - b;
            //取余的位置
            a %= 1000000007;
        }
        return a;
    }
}
</code></pre>

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